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We have a metric on $\mathbb{R}^2$ defined as: $d(x,y) = \max(|x_1-y_1|,|x_2-y_2|)$ where $x = (x_1,x_2)$ and $y = (y_1,y_2)$. To satisfy the triangle inequality, we must show that $\max(|x_1-y_1|,|x_2-y_2|) \leq\max(|x_1-z_1|,|x_2-z_2|) +\max(|z_1-y_1|,|z_2-y_2|)$. I am having trouble formally proving this with the $max$ statements in there without hand waving. We know from properties of the absolute value that $|x+y| \leq |x| + |y|$. Can anyone show me a formal way to do this?

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  • $\begingroup$ It's not clear yet what you're asking. $\endgroup$ Feb 17, 2015 at 9:07

3 Answers 3

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Cases:

  1. $|x_1-z_1|<|x_2-z_2|$ and $|z_1-y_1|<|z_2-y_2|$ (or both are reversed): then by your assumed triangle inequality $|x_1-y_1|<|x_2-y_2|$ (resp. reversed) and the rest is easy.
  2. the remaining case, $|x_1-z_1|<|x_2-z_2|$ but $|z_1-y_1|>|z_2-y_2|$ (or if both are reversed, this goes through the same way m.m.): Consider $\max(|x1-y1|,|x2-y2|)$, WLOG $|x2-y2|$. By your assumed triangle inequality it's $<|x_2-z_2|+|z_2-y_2|$, but that's $<|x_2-z_2|+|z_1-y_1|=\max(|x_1-z_1|,|x_2-z_2|)+\max(|z_1-y_1|,|z_2-y_2|)$ qed.
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$|x_i-y_i|\leq |x_i-z_i|+|z_i-y_i|\leq \max (\{|x_i-z_i|:i\in\{1, 2\}\})+\max (\{|z_i-y_i|:i\in\{1, 2\}\})= d (x, y)+d (y, z) $

Note that this can be easily generalised to higher dimensions.

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You're really asking a simpler question.

Let $$ x_1 \le x_2 + x_3\\ y_1 \le y_2 + y_3 \\ $$ where all these numbers are positive.

You want to show that $$ max\{x_1,y_1\}\le max\{x_2,y_2\}+max\{x_3,y_3\} $$ Consider the right side of the inequality. The sum of the two maxima must be at least $x_2+x_3$; it could well be strictly larger Likewise, that sum must be at least $y_2+y_3$. But $x_1 \le x_2+x_3$, and also $y_1 \le y_2+y_3$. Thus the maximum of $x_1$ and $y_1$ is most assuredly less than the right side of that inequality.

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